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Homework Help: Do I have a problem? Basic proportion problem.

  1. Aug 11, 2012 #1
    Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

    The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete. What is the total number of questions on his assignment? The answer was 50.

    I got that by

    [itex]\frac{7}{10}[/itex]=[itex]\frac{(x-15)}{x}[/itex] Followed by cross multiplying and dividing.

    However the book does it simply by doing the subtraction 10-7 = 3 then setting it up as

    [itex]\frac{3}{10}[/itex]=[itex]\frac{15}{x}[/itex] Then cross multiply and divide.

    I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a highschool record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 11, 2012 #2
  4. Aug 11, 2012 #3


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    Science Advisor

    Do you meant "15 more questions to complete"? What you wrote could be interpreted as meaning has completed 15 questions. If he has already completed 7/10 of his assignment, he still has 3/10 left. If he has 15 questions more to complete, letting "x" be the total number of problems, (3/10)x= 15 so x= (10/3)(15)= 150/3= 50.

    Okay, that is, they are calculating 1- 7/10= 10/10- 7/10= (10- 7)/10= 3/10 as I did.

    As long as you have linear problems, you can set them up as "proportions" and get the correct answer.

  5. Aug 11, 2012 #4
    Thanks guys
    Last edited: Aug 11, 2012
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